Richard Price
2007-Aug-14 14:36 UTC
[R] Comparing long species lists via Sorensons dissimilarity
I have 4 very large species lists and I would like to compare them.
I have the following results from running Sorenson’s dissimilarity tests:
Norfolk Fens compared to Suffolk Coastal Fens:
QS=0.583961142689298
Norfolk Fens compared to Breckland Edge Fens:
QS=0.714896020281379
Norfolk Fens compared to Other Fens:
QS=0.78572348898302
Suffolk Coastal Fens compared to Breckland Edge Fens: QS=0.78572348898302
Suffolk Coastal Fens compared to Other Fens:
QS=0.855011155835705
Breckland Edge Fens compared to Other Fens:
QS=0.175091076893185
Does anyone know how to interpret the above results? I have read a paper that
states that when comparing two samples if QS is less than 50% than the samples
are considered dissimilar, what is 100%? I mean 50% of what?
There are a number of other tests that I can run using R. These are squared
euclidan, manhattan, bray-curtis, jaccard, ruzicka, (dis)similarity ratio,
ochiai, cosine compliment, and Raup-crick. Would it be advantageous to run these
now that I have my sorensons result? It is especially easy for me to run the
Bray-Curtis all I would need to do is change the terms from ‘binary’ to
‘minimum’ and re-run. Would Bray-Curtis be a Percentage Similarity unlike the
others?
Here is an example of what happens when I run sorensons:
#Compare Breckland with Other
> a=table(breckland)
> J=sum(breckland*other)
> A=sum(breckland^2)
> B=sum(other^2)
> brecklandOther <- designdist(a,method=(A+B-2*J)/(A+B), terms =
c("binary"))
[1] 0.1750911
attr(,"call")
designdist(x = a, method = (A + B - 2 * J)/(A + B), terms =
c("binary"))
attr(,"method")
[1] "0.175091076893185 binary"
I do not know why the following lines are included (perhaps R is trying to
calculate the header row? But it doesn’t seem to stop the result getting out.
attr(,"call")
designdist(x = a, method = (A + B - 2 * J)/(A + B), terms =
c("binary"))
attr(,"method")
I’ve also managed to get diversity indices from R but someone told me that
comparing diversity indices is not good because of the way that they are
calculated.
>norfolkShannon
3.653768
> suffolkShannon
Suffolk
3.646138
> brecklandShannon
Breckland
4.051742
> otherShannon
Other
3.587403
For the above statistical methods is there a way of laying out the data
perhaps via a graph that will show how similar/dissimilar the sites are?
Many thanks for yesterdays help and in advance for any help given today.
Richard Price
University of Birmingham.
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